When does a null integral implies that a form is exact? It is trivial to prove that the integration of a $(n-1)$ exact form on the boundary of a $n$-manifold is 0. What about the contraposative ? If the integration of a $(n-1)$-form on the boundary of a $n$-manifold is 0, is this form exact ? If not, are there particular conditions to satisfy for this to be the case ?
For example, in the case of $S^1$, any $1$-form can be written as $f(\theta)d\theta=c d\theta+dg(\theta)$, $c$ being the integral around $S^1$, and $g$ a differentiable function on $S^1$. Now in this case if the integral is 0, it implies that the form is exact. I wondered in the general case if such decomposition is always possible, because if so, it can be proven that the integral being 0 implies that the form is exact. Is this correct ?
 A: Stokes' theorem works both ways, but you have to be careful. In particular, a $n$-form $\omega$ is closed if and only if its integral over every $n$-boundaries is zero, not just one particular!
For every $(n+1)$-dimensional region $\Omega$ we have:
$$
\int_{\partial\Omega}\omega=\int_\Omega d\omega.
$$
If the expression on the left is zero for every $\Omega$, so is the one on the right, and it means that $d\omega=0$.
If we want $\omega$ to be exact, then we don't have to limit ourselves to $n$-boundaries, but we want its integral to be zero on every closed region!
That is, any region $A$ with $\partial A=0$.
This because what cause a closed form not to be exact are the very "holes".
I don't know if from this requirement you can explicitly find that $\omega=d\eta$ for some $\eta$ (that is, I don't know if this reasoning provides you such an $\eta$). 
What is immediate to check (just use Stokes), though, is that exact forms yield a zero integral over general closed regions, not just boundaries.
A: I think the result you are looking for is this: Top deRham cohomology group of a compact orientable manifold is 1-dimensional, with the isomorphism $H^n_{dR}(M^n)\to \mathbb R$ given by $\omega\mapsto \int_M \omega$. (One should add "connected" to the assumptions.) In your question, $n$ is actually $n-1$. 
The proof of $H^n_{dR}(M^n)\approx \mathbb R$  is not hard but lengthy: see section 8.1, titled de Rham cohomology in the top dimension, of Nigel Hitchin's notes.
See also Poincaré Duality with de Rham Cohomology. And for the non-compact case, De Rham cohomology for non-compact manifolds.  
A: No, and my limited knowledge of the topic is that it depends heavily on the topology of a space. For instance, in a multiply connected 2-manifold, there are closed forms which are not exact (they are not the boundary of any submanifold, since they "go around they hole").
